Below are a few sample problems. You are responsible for all material covered in lecture and homework assignments. DO NOT ASSUME that a subject not covered in the sample exam below will not show up on the exam!
During the exam, you will be provided with an equation sheet. You will be allowed to use only the equations on the sheet, or others that you derive from those on the sheet. You may download the Equation sheet. Resist the temptation to simply look for a similar problem in the text or notes (the standard "get the homework done" technique). That will better prepare you for the exam. If you can't solve the problems, come see me. And don't wait until Wednesday afternoon!!
I've haven't double check my arithmetic, so send me an email if you spot some mistakes.
The x and y components of the displacement is Dx = 100 + 0+ 100 cos45o = 171 m, and Dy = 0 + 200 + 100sin45o = 271 m. Magnitude is sqrt(1712 + 2712) = 320 m at q = tan-1(271/171) = 58o north of east.
The total distance is 120 mph x 1/3 hr + 100 mph x 2/5 hr = 40 + 40 = 80 mi, traveled in total time 20 + 24 = 44 minutes = 0.73 hr. So ave speed = 80.0 mi/0.73 hr = 109 mph.
For average velocity, we need the x and y components (Dx, Dy) of the displacement over total time. v = D/Dt = (40, 40) mi /0.73 hr = (55, 55) mph. That is, vx = 55 mph and vy = 55 mph. (Or magnitude 77 mph at 45o north of east.)
Use the convenience equation 302 = 02 + 2a(12 ft) --> a = 37.5 ft/s2. F = ma. So F = (170/32) x 37.5 = 200 lb. The last question must be divided into to parts. During the acc part, 12 = 1/2(37.5)t2 and during the remaining 80-12 = 68 = 30t, so total time = 0.64 + 2.27 = 2.91 sec.
Use symmetry, so it takes 3 seconds to go up and the velocity must -12 m/s when it arrives back at your hand. So -12 = +12 - g(6s) --> g = 4 m/s2. To get the height. y = 12(3)-1/2(4)(32) = 18 m.
Captain Bligh wishes to put a hole at water level in the good ship Lollipop with his cannon. The cannon is 16 ft above the water and fires horizontally. How much gun powder (1 gm for every 1.0 ft/s velocity) does he need if the Lollipop is 300 ft away?x = 300 = voxt. voy = 0 so we get t from 0 = 16-1/2(32)t2, so vox = 300/1.0 = 300 ft/s so that's 0.30 kg of powder.
vox = 50 cos37o = 40 ft/s and voy = 50 sin37o = 30 ft/s. 0 = 6 + 30t - 1/2(32)t2 --> t = 2.06 s, so x = 40(2.06) = 82 ft.
F = (4.0 x 104, 2.0 x 104) = 10,000a. So a = (4.0, 2.0) m/s2. Magnitude is sqrt(4.02 + 2.02) = 4.47 m/s2 at q = tan-1(2/4) = 27o north of east.